Polynomial division & factor theorem

For a polynomial \(P\) and \(a\ne0\):
<div class="content-box-placement content-box-placement--full content-box-placement--align-center" style="display:flex;justify-content:stretch;max-width:100%"><div class="content-box content-box--emphasis" data-content-box-preset="emphasis" style="box-sizing:border-box;border-color:var(--content-box-border, #1E6FBA);border-style:solid;border-width:0px;border-radius:0px;background-color:var(--content-box-background, #F5F8FC);color:var(--content-box-foreground, #111827);padding:12px 16px;font-family:'EB Garamond', serif;font-size:16px;line-height:1.35;width:100%;max-width:100%;border-left-width:3px">\(ax+b\text{ is a factor of}\hspace{0.5em}\mathrm{P}(x)\quad\Longleftrightarrow\quad \mathrm{P}\!\left(-\dfrac ba\right)=0\)</div></div>
If&nbsp;\(r\ne s\), then&nbsp;\((x-r)(x-s)\) is a factor exactly when&nbsp;\(\mathrm{P}(r)=\mathrm{P}(s)=0\).
For a repeated factor, testing the root once is insufficient; divide again to check the required multiplicity.
Example
Given that&nbsp;\(x-2\) and \(x+1\) are factors of \(\mathrm{P}(x)=x^3+ax^2+bx-6\), find \(a\) and \(b\).
\(\begin{aligned} \mathrm{P}(2)=8+4a+2b-6&=0 \\ 2a+b&=-1 \end{aligned}\)
\(\begin{aligned} \mathrm{P}(-1)=-1+a-b-6&=0 \\ a-b&=7 \end{aligned}\)
\(a=2,\ b=-5\)
Check: \(x^3+2x^2-5x-6=(x-2)(x+1)(x+3)\).

\(ax+b\text{ is a factor of }\mathrm{P}(x)\quad\Longleftrightarrow\quad \mathrm{P}\!\left(-\dfrac ba\right)=0\)

If \(r\ne s\), then \((x-r)(x-s)\) is a factor exactly when \(\mathrm{P}(r)=\mathrm{P}(s)=0\).

For a repeated factor, testing the root once is insufficient; divide again to check the required multiplicity.

Given that \(x-2\) and \(x+1\) are factors of \(\mathrm{P}(x)=x^3+ax^2+bx-6\), find \(a\) and \(b\).

\(\begin{aligned} \mathrm{P}(2)=8+4a+2b-6&=0 \\ 2a+b&=-1 \end{aligned}\)

\(\begin{aligned} \mathrm{P}(-1)=-1+a-b-6&=0 \\ a-b&=7 \end{aligned}\)

Check: \(x^3+2x^2-5x-6=(x-2)(x+1)(x+3)\).

Factor Theorems

Factor Theorems

Video Solution:

Solution:

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